How to calculate this angular velocity problem

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Homework Help Overview

The problem involves a turntable with a specified mass and diameter, rotating at a given angular velocity. Two blocks fall onto the turntable and stick, prompting a discussion about the calculation of the new angular velocity after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of angular momentum and moment of inertia, with some questioning the correct formulas to use for the blocks and the turntable. There is also an exploration of the contributions of different masses to the overall moment of inertia.

Discussion Status

Participants are actively engaging with the problem, with some providing calculations and others seeking clarification on the formulas. There is a recognition of the need to combine the moments of inertia of the turntable and the blocks, indicating a productive direction in the discussion.

Contextual Notes

There is mention of the turntable being on frictionless bearings and the specific masses and dimensions involved, which may influence the calculations. Some participants express uncertainty about their approaches and results.

karzy
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A 2.3kg , 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500g blocks fall from above, hit the turntable simultaneously at opposite ends of a diagonal, and stick.
What is the turntable's angular velocity, in rpm, just after this event?I first cacluted the angular momentum as .120428 using I*W so I then set that equal to I*W where I=.5(3.3)(.1)^2
and w is 2pi(frequency) and tried solving for frequency. and converting that back to rpm. But 69.7 rpm wasn't correct so I'm missing something or approaching this wrong!

Any suggestions?
 
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What's the new moment of inertial after the blocks stick to the turntable?

Chet
 
i used the formula (1/2)(M)R^2 which I used 3.3 for the mass and .1 for the R which gave me .0165.
 
karzy said:
i used the formula (1/2)(M)R^2 which I used 3.3 for the mass and .1 for the R which gave me .0165.
This formula applies to the disk, but not to the blocks. The blocks are at the outside radius, so their contributions to the moment of inertial (per unit mass) are greater. Do you know the formula for the moment of inertial of a mass at a distance r from the center of rotation?

Chet
 
I=MR^2 where m is the mass and r is distance from the center?
 
Ahhh i just got it! I needed to add the three Ineritas together then solve for the velocity.Thanks for the help!
 

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